1.

A particle accelerates from rest at a constant rate for some time and attains a velocity of 8m/s. Afterwards it deaccelerates with the constant rate and comes to rest .If the total time taken is 4sec, what is the distance travelled?​

Answer»

GIVEN :

  • Initial VELOCITY, u = 0 m/s.
  • Final velocity, v = 8 m/s.
  • time, t = 4 sec.

TO FIND :

  • Distance travelled = ?

CONCEPTS USED :

  • Newton's first law of motion.
  • Newton's third law of motion.

SOLUTION :

First we have to find the acceleration (a) = ?

USING Newton's first law of motion,

V = u + at

=> 8 = 0 + a × 4

=> 8 = 4a

=> a = 8/4

\qquad \large {\boxed {\sf a \ = \ 2 \ m/s^{2}}}

Now,

To find the distance travelled (S) = ?

Using the Newton's third law of motion,

V² = u² + 2as

=> 8² = 0² + 2 × 2 × S

=> 64 = 0 + 4 × S

=> 64 = 4 S

=> S = 64/4

\qquad \large {\boxed {\sf Distance \ = \ 16 \ m}}

•°• The distance travelled is 16 METERS.

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★ ALSO KEEP IN MIND :

  • Newton's first law of motion : V = u +at.
  • Newton's second law of motion : F = ma.
  • Newton's third law of motion : V² = u² + 2as.


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